Question: Solve for $q$, $ \dfrac{5}{2q^2} = -\dfrac{1}{q^2} - \dfrac{q - 7}{5q^2} $
Answer: First we need to find a common denominator for all the expressions. This means finding the least common multiple of $2q^2$ $q^2$ and $5q^2$ The common denominator is $10q^2$ To get $10q^2$ in the denominator of the first term, multiply it by $\frac{5}{5}$ $ \dfrac{5}{2q^2} \times \dfrac{5}{5} = \dfrac{25}{10q^2} $ To get $10q^2$ in the denominator of the second term, multiply it by $\frac{10}{10}$ $ -\dfrac{1}{q^2} \times \dfrac{10}{10} = -\dfrac{10}{10q^2} $ To get $10q^2$ in the denominator of the third term, multiply it by $\frac{2}{2}$ $ -\dfrac{q - 7}{5q^2} \times \dfrac{2}{2} = -\dfrac{2q - 14}{10q^2} $ This give us: $ \dfrac{25}{10q^2} = -\dfrac{10}{10q^2} - \dfrac{2q - 14}{10q^2} $ If we multiply both sides of the equation by $10q^2$ , we get: $ 25 = -10 - 2q + 14$ $ 25 = -2q + 4$ $ 21 = -2q $ $ q = -\dfrac{21}{2}$